Answer
$I_{O}=6.23$ kg*m$^2$
Work Step by Step
$m_{square}=(1.40)(1.40)(0.05)(50)=4.19$ kg
$m_{circle}=(\pi)(0.15)^2(0.05)(50)=1.77$ kg
$I_{O}=\Sigma I_{G}+md^2=(\frac{1}{12})(4.9)(2)(1.40)^2-(\frac{1}{2})(0.177)(0.15)^2+(4.9-0.177)(1.4sin(45[{\deg}]))$
$=6.23$ kg*m$^2$
